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2 years ago

Question :

2 answers:

riadik2000 [5.3K]2 years ago

8 0

The second law of motion states that: the acceleration of an object is dependent upon two variables: - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object

<h3>Meaning of Motion</h3>

Motion can be defined as the process of changing position willingly or due to a force applied.

Motion can be seen in different forms and types depending on the object.

In conclusion, The second law of motion is used to deduce the formula for acceleration.

Learn more about second law of motion: brainly.com/question/2009830

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snow_lady [41]2 years ago

4 0

Answer:

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Explanation:

Hope it's helpful to you

You might be interested in

How is alternating current generated?

Viefleur [7K]

AC can be produced using a device called an alternator. This device is a special type of electrical generator designed to produce alternating current. A loop of wire is spun inside of a magnetic field, which induces a current along the wire.

3 0

3 years ago

Select. The star cycle that is accurate

melomori [17]

One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.

<h3 /><h3>What is the solar system?</h3>

The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.

The complete question is;

"Choose the correct star life cycle.

A. Supernova, star as well as red giant, the nebula is incorrect.

B. Nebula, white dwarf, planetary nebula That is incorrect.

C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!

D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."

Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.

Hence option C is correct.

To learn more about the solar system refer to the link;

brainly.com/question/1207587

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5 0

2 years ago

A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig

olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

$speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s$

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

$y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\$

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

$6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs$

the horizontal distance covered at this speed is

$y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m$

4 0

3 years ago

You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The freque

tiny-mole [99]

The frequency of the other wave is 613 Hz or 607 Hz.

The difference between the frequencies of two waves is called the beat frequency.

Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.

Which has a higher frequency is not mentioned. Therefore, there are two possibilities.

Δf = | 610 - 613 | = 3

Δf = | 610 - 607 | = 3

Therefore, the frequency of the other wave is 613 Hz or 607 Hz.

Learn more about beat frequency here:

brainly.com/question/14157895

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7 0

2 years ago

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago